Problemas4eso.pdf

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Problemas de 4

o

ESO

Isaac Musat Herv´as

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´

_{Indice General}

1 Problemas de ´Algebra 7

1.1 N´umeros Reales . . . 7

1.1.1 Los n´umeros . . . 7

1.1.2 Intervalos . . . 14

1.1.3 Ecuaciones Bicuadradas . . . 18

1.2 N´umeros Racionales . . . 19

1.2.1 Operaciones con n´umeros racionales . . . 19

1.2.2 Ecuaciones Racionales . . . 22

1.3 Logaritmos . . . 26

1.3.1 Ecuaciones Logar´ıtmicas . . . 26

1.3.2 Sistemas de Ecuaciones Logar´ıtmicas . . . 34

1.4 Exponenciales . . . 42

1.4.1 Ecuaciones Exponenciales: . . . 42

1.4.2 Sistemas de Ecuaciones Exponenciales: . . . 47

1.5 Ecuaciones Logar´ıtmicas y Exponenciales . . . 53

1.6 Sistemas de Ecuaciones no Lineales . . . 60

1.7 Inecuaciones . . . 60

1.7.1 Inecuaciones . . . 60

1.7.2 Sistemas de Inecuaciones . . . 80

1.8 Polinomios . . . 84

1.8.1 Introducci´on . . . 84

1.8.2 Teorema del Resto . . . 89

1.8.3 Descomposici´on Polin´omica . . . 90

1.8.4 Simplificaci´on . . . 94

1.8.5 Máximo Común Divisor y M´ınimo Común Múltiplo . 95 1.8.6 Simplificación de expresiones racionales de polinomios 101 1.8.7 Ecuaciones Polinómicas . . . 105

2 Problemas de Geometr´ıa 109 2.1 Trigonometr´ıa . . . 109

2.1.1 Angulos . . . 109´

2.1.2 Razones Trigonom´etricas . . . 114

2.1.3 Resoluci´on de Tri´angulos . . . 123

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2.1.4 Aplicaciones . . . 126

2.2 Vectores . . . 139

2.2.1 Operciones con Vectores . . . 139

2.2.2 Distancia entre dos puntos . . . 140

2.2.3 Divisi´on de un segmento . . . 140

2.2.4 Punto medio y sim´etrico . . . 143

2.2.5 Angulo entre dos vectores . . . 144´

2.2.6 Varios . . . 145

2.3 Geometr´ıa Anal´ıtica . . . 150

2.3.1 Ecuaciones de la Recta . . . 150

2.3.2 Intersecci´on de dos rectas . . . 154

2.3.3 Distancias . . . 156

2.3.4 Angulos . . . 159´

2.4 C´onicas . . . 160

2.4.1 Circunferencia . . . 160

2.4.2 Elipse . . . 164

2.4.3 Hip´erbola . . . 165

3 Problemas de An´alisis 167 3.1 Sucesiones . . . 167

3.1.1 T´erminos de una sucesi´on . . . 167

3.1.2 Sucesiones crecientes y acotadas: . . . 169

3.1.3 Progresiones aritm´eticas . . . 171

3.1.4 Progresiones geom´etricas . . . 179

3.2 L´ımites de sucesiones . . . 187

3.2.1 Idea intuitiva . . . 187

3.2.2 Definici´on . . . 189

3.2.3 Sucesiones que tienden a infinito . . . 191

3.2.4 C´alculo de L´ımites de sucesiones . . . 192

3.2.5 N´umero e . . . 195

3.2.6 Varios . . . 197

3.3 Funciones . . . 199

3.3.1 Concepto de funci´on, Dominio y Recorrido . . . 199

3.3.2 Funciones definidas a trozos . . . 202

3.3.3 Crecimiento y decrecimiento. M´aximos y m´ınimos . . 202

3.3.4 Funciones acotadas. Funciones sim´etricas. Estudio gr´afico de la continuidad. Puntos de corte con los ejes. 205 3.3.5 Operaciones con funciones. Funciones rec´ıprocas . . . 208

3.3.6 Puntos de Corte . . . 212

3.3.7 Simetr´ıa . . . 214

3.3.8 Composici´on de Funciones . . . 220

3.3.9 Funci´on Inversa . . . 223

3.3.10 Monoton´ıa . . . 225

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´_{INDICE GENERAL} ₅

3.4.1 L´ımite de una funci´on en un punto . . . 226

3.4.2 L´ımite de una funci´on en el infinito . . . 228

3.4.3 C´alculo de l´ımites de funciones racionales . . . 230

3.5 Continuidad . . . 233

3.5.1 Continuidad en un punto y en un intervalo . . . 233

3.5.2 Tipos de discontinuidad . . . 237

3.5.3 Continuidad y Operaciones: . . . 240

3.5.4 Problemas de Continuidad . . . 242

3.6 As´ıntotas de una funci´on . . . 256

3.7 Problemas de L´ımites . . . 260

3.8 Problemas Varios . . . 290

3.8.1 Problemas de Dominio . . . 290

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Cap´ıtulo 1

Problemas de ´

Algebra

1.1 N´

umeros Reales

1.1.1 Los n´umeros

Problema 1 Indica el conjunto más pequeño al que pertenece cada uno de los siguientes números:

−7 ; 12 ; 0 ; π ; 2,333. . . ; −3₇ ; 2,1010010001. . .

Soluci´on:

−7 es un número entero −7∈Z. 12 es un número natural 12∈N. 0 es un número natural 0∈N.

π es un n´umero irracional.

2,333. . .es un n´umero racional 2,b3∈Q.

−3

7 es un n´umero racional − 3 7 ∈Q.

2,1010010001. . .es un n´umero irracional.

−12 ; 5₂ ; √7 ; 23 ; 7,34 ; 5,222272727. . . ; 3,7770700700070000. . .

Soluci´on:

−12 es un n´umero entero −12∈Z.

5

2 es un n´umero racional 5 2 ∈Q.

√

7 es un número irracional. 23 es un número natural 23∈N. 7,34 es un número racional 7,34∈Q.

5,222272727. . .es un n´umero racional5,2222c72∈Q.

3,7770700700070000. . . es un n´umero irracional.

3 ; −2 ; −4₃ ; 4,3327832783278. . . ; 4,33133113331113333. . . ; √7 ;

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π; 7,1203870387. . .; 2+₂ 5

Soluci´on:

• 3 es un n´umero natural 3∈N.

• −2 es un n´umero entero−2∈Z.

• −4

3 es un n´umero racional− 4 3 ∈Q.

• 4,3327832783278. . . es un n´umero racional 4,33278d ∈Q.

• 4,33133113331113333. . .es un n´umero irracional.

• √7 es un n´umero irracional.

• π es un n´umero irracional.

• 7,1203870387. . .es un n´umero racional 7,120387d ∈Q

• 2+

√

5

2 es un n´umero irracional.

−3 ; 2 ; −4

3 ; 4,3322277722227777. . . ; 4,33278278278. . . ;

√

5 ;

π; 2+ √

3

2 ; 7,1203870387. . .

Soluci´on:

• −3 es un n´umero entero 3∈Z.

• −4

3 es un n´umero racional− 4 3 ∈Q.

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1.1. N ´UMEROS REALES 9

• 2+

√

3

3 4 ;

√

2 ; 5 ; 0,12348348. . .; 0,123123412345. . .; −3 ; π; 0,110011100011110000. . .; 0; 2₅.

Soluci´on:

• 3₄ es un n´umero racional 3₄ ∈Q.

• 0,12348348. . . es un n´umero racional 0,12348d ∈Q.

• 0,123123412345. . . es un n´umero irracional.

• −3 es un n´umero entero −3∈Z.

• 2₅ es un n´umero racional 2₅ ∈Q

1 4 ;

√

3 ; 7 ; 0,12359359. . .; 0,123123412345. . .; −2 ; π; 0,110011100011110000. . .; 0; 4₅.

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• 1₄ es un n´umero racional 1₄ ∈Q.

• 0,12359359. . . es un n´umero racional 0,12359d∈Q.

• 4₅ es un n´umero racional 4₅ ∈Q

3 ; −√5 ; 2,125125125. . . ; −9

4 ; −1

Soluci´on:

3 es un n´umero natural 3∈N.

−√5 es un n´umero irracional.

2,125125125. . . es un n´umero racional 2,125d∈Q.

−9

−1 es un n´umero entero −1∈Z.

−3 ; 0,56 ; 0 ; π ; 1,1122111222. . . ; −3₄ ; 2; 7,161616. . .; 3,21213214215. . .; 8,666. . .

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−3 es un número entero −3∈Z. 0,56 es un número racional 0,56∈Q. 0 es un número natural 0∈N.

π es un n´umero irracional.

1,1122111222. . .es un n´umero irrracional.

−3

7,161616. . . es un n´umero racional 7,16c∈Q.

3,21213214215. . . es un n´umero irracional. 8,666. . .es un n´umero racional 8,b6∈Q

−1 ; 0,71 ; 0 ; √2 ; 1,1133111333. . .; −1₇ ; 2; 9,262626. . .; 3,21213214215. . .; 3,333. . .

Soluci´on:

−1 es un número entero −1∈Z. 0,71 es un número racional 0,71∈Q. 0 es un número natural 0∈N.

√

1,1133111333. . .es un n´umero irrracional.

−1

9,262626. . . es un n´umero racional 9,26c∈Q.

3,21213214215. . . es un n´umero irracional. 3,333. . .es un n´umero racional 3,b3∈Q

2 ; −3 ; 3₄ ; 3,7728122812. . .; 5,1133111333. . .; √3 ; π; 3,230173017. . .;

1−√5 2 ; 0

Soluci´on:

• 3

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• 5,1133111333. . . .es un n´umero irracional.

• 3,230173017. . . es un n´umero racional 3,23017d ∈Q

• 1−√5

2 ; −3 ; 3₄; 3,7728122812. . .; 5,1133111333. . .; √3 ; π; 3,230173017. . .;

1−√5 2 ; 0

Soluci´on:

• 3

• 3,7728122812. . .es un n´umero racional 3,772812d ∈Q.

• 3,230173017. . . es un n´umero racional 3,23017d ∈Q

• 1−√5

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3 ; −2 ; 1₄ ; 2,7728122812. . .; 6,1133111333. . .; √5 ; π; 4,230273027. . .;

1−√5 2 ; 0

Soluci´on:

• 1

• 2,7728122812. . .es un n´umero racional 2,772812d ∈Q.

• 1−√5

−3 ; 2,71 ; 0 ; √5 ; 1,2233222333. . .; −13₇ ; 5; 11,163636. . .; 4,21132142152. . .; 5,333. . .

Soluci´on:

−3∈Z; 2,71∈Q; 0∈N; √5∈irracional; 1,2233222333. . .∈irracional; −13

7 ∈

Q; 5∈N; 11,163636. . .∈Q;

4,21132142152. . .∈irracional; 5,333. . .∈Q

3 ; 2,7171. . .; π ; √9 ; 3,22442224444. . . ; −7

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7,2122132142. . .; 6,111. . .

Soluci´on:

3∈N; 2,7171. . .∈Q; π∈irracional; √9 = 3∈N ; 3,22442224444. . .∈

irracional ; −7

9 ∈Q; 0∈N; 23,163737. . .∈Q; 7,2122132142. . .∈irracional; 6,111. . .∈Q

5; 4,8282; 1 +

√

5 2 ;

√

81 ; 3,2277222777. . .; −5

9; 21,253838. . .; 7,112113114. . .; 4,111. . .

Soluci´on:

5∈R; 4,8282. . .∈Q; 1 +

√

5

2 ∈irracinal

√

81 = 9∈R; 3,2277222777. . .∈

irracinal; −5

9 ∈Q; 0∈N; 21,253838. . .∈Q; 7,112113114. . .∈irracinal; 4,111. . .∈Q

6∈N; 7,5252. . .∈Q; π∈irracionales ; √36 = 6∈N; 3,5577555777. . .∈

irracionales ; −3

4 ∈Q ; −1∈Z; 1,143939. . .∈Q; 7,772773774. . .∈irracionales; 9,999. . .∈Q

1.1.2 Intervalos

Problema 17 Dibuja los siguientes intervalos en la recta real:

1. |x−3|<1

2. |x−5| ≤3

(Recuerda la definici´on de entorno, E(a, r) ={x∈R:|x−a|< r}. Soluci´on:

1. E(3,1) ={x∈R:|x−3|<1}=⇒E(3,1) = (3−1,3 + 1) = (2,4).

2. E[5,3] ={x∈R:|x−5| ≤3}=⇒E[5,3] = [5−3,5 + 3] = [2,8].

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2. |x−1| ≤3

1. E(4,2) ={x∈R:|x−4|<2}=⇒E(4,2) = (4−2,4 + 2) = (2,6).

2. E[1,3] ={x∈R:|x−1| ≤3}=⇒E[1,3] = [1−3,1 + 3] = [−2,4].

1. {x∈R:−3≤x <7}

2. {x∈R: 4< x <8}

3. {x∈R:x≥3}

4. {x∈R:x <−1}

5. {x∈R:|x−3| ≤5}

6. {x∈R:|x+ 1|<2}

1. {x∈R:−3≤x <7}= [−3,7)

2. {x∈R: 4< x <8}= (4,8)

3. {x∈R:x≥3}= [3,+∞)

4. {x∈R:x <−1}= (−∞,−1)

5. {x∈R:|x−3| ≤5}= [3−5,3 + 5] = [−2,8]

6. {x∈R:|x+ 1|<2}= (−1−2,−1 + 2) = (−3,1)

1. {x∈R:−2≤x <6}

2. {x∈R: 1< x <9}

3. {x∈R:x≥1}

4. {x∈R:x <−3}

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6. {x∈R:|x+ 1|<3}

1. {x∈R:−2≤x <6}= [−2,6)

2. {x∈R: 1< x <9}= (1,9)

3. {x∈R:x≥1}= [1,+∞)

4. {x∈R:x <−3}= (−∞,−3)

5. {x∈R:|x−2| ≤5}= [2−5,2 + 5] = [−3,7]

6. {x∈R:|x+ 1|<3}= (−1−3,−1 + 3) = (−4,2)

Problema 21 Dados los intevalos A = (−1,4] B = (−∞,2] y C = (1,3), calcularA∩B,A∪C,B∩C yB∪C

Soluci´on:

A∩B = (−1,2], A∪C= (−1,4], B∩C = (1,2], B∪C = (−∞,3)

Problema 22 Escribe de todas las maneras que conozcas los siguientes intervalos

1. {x∈R:|x−2| ≤8}

2. {x∈R:|x+ 1|<9}

1. {x∈R:|x−2| ≤8}=E(2,8) ={x∈R:−6≤x≤10}= [−6,10]

2. {x∈R:|x+ 1|<9}=E(−1,9) ={x∈R:−10< x <8}= (−10,8)

Soluci´on:

A∩B = (−2,2], A∪C = (−2,4], B∩C= (1,2], B∪C= (−∞,4)

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1. {x∈R:|x−5| ≤5}

2. {x∈R:|x+ 2|<8}

1. {x∈R:|x−5| ≤5}=E(5,5) = [0,10] ={x∈R: 0≤x≤10}

2. {x∈R:|x+ 2|<8}=E(−2,8) = (−10,6) ={x∈R:−10< x <6}

Problema 25 Dados los intevalos A = (−3,4] B = (−3,2] y C = (0,4], calcularA∩B,A∪C,B∩C yB∪C

Soluci´on:

A∩B= (−3,2], A∪C= (−3,4], B∩C= (0,2], B∪C= (−∞,4)

1. {x∈R:|x−1| ≤7}

2. {x∈R:|x+ 4|<10}

1. {x∈R:|x−1| ≤7}=E(1,7) = [−6,8] ={x∈R:−6≤x≤8}

2. {x∈R:|x+ 4|<10}=E(−4,10) = (−14,6) =

={x∈R:−14< x <6}

Soluci´on:

A∩B= (−3,3], A∪C= (−3,7], B∩C= (0,3], B∪C= (−∞,7)

1. {x∈R:|x−2| ≤12}

2. {x∈R:|x+ 3|<11}

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1. {x∈R:|x−2| ≤12}=E(2,12) = [−10,14] =

={x∈R:−10≤x≤14}

2. {x∈R:|x+ 3|<11}=E(−3,11) = (−14,8) =

{x∈R:−14< x <8}

1.1.3 Ecuaciones Bicuadradas Problema 29

x4−8x2−9 = 0

Soluci´on:

Hacemos z=x2 ₌_⇒_z2₋₈_z₋_{9 = 0 =}_⇒ _z_{= 9} _y _z₌₋₁_.

z= 9 =x2 =⇒x=±3

z=−1 =x2 No Vale

Problema 30

x4−14x2−32 = 0

Soluci´on:

Hacemos z=x2 =⇒z2−14z−32 = 0 =⇒ z= 16y z=−2.

z= 16 =x2 =⇒x=±4

z=−2 =x2 No Vale

Problema 31

x4−80x2−81 = 0

Soluci´on:

Hacemos z=x2 =⇒z2−80z−81 = 0 =⇒ z= 81y z=−1.

z= 81 =x2 =⇒x=±9

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1.2. N ´UMEROS RACIONALES 19

Problema 32

x4−2x2−8 = 0

Soluci´on:

Hacemos z=x2=⇒z2−2z−8 = 0 =⇒ z= 4 y z=−2.

z= 4 =x2 =⇒x=±2

z=−1 =x2 No Vale

Problema 33

x4−24x2−25 = 0

Soluci´on:

Hacemos z=x2=⇒z2−24z−25 = 0 =⇒ z= 25 yz=−1.

z= 25 =x2 =⇒x=±5

z=−1 =x2 No Vale

Problema 34

x4+x2−20 = 0

Soluci´on:

Hacemos z=x2=⇒z2+z−20 = 0 =⇒ z= 4 y z=−5.

z= 4 =x2 =⇒x=±2

z=−5 =x2 No Vale

1.2 N´

umeros Racionales

1.2.1 Operaciones con n´umeros racionales

Problema 35 Racionalizar las siguientes expresiones:

1. √2

5; 3

√

3

2. 1 1 +√5;

2 5−√5

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1. √2

5 = 2√5

5

3

√

3 = 3√3

3 =

√

3

2. 1 1 +√5 =

1−√5

(1 +√5)(1−√5) =

1−√5 (1−(√5)2₎ =

1−√5 1−5 =−

1−√5 4

2 5−√5 =

2(5 +√5) (5−√5)(5 +√5) =

2(5 +√5) (52₋₍√₅₎2₎ =

5 +√5 10

1. √3

7; 5

√

5

2. 1 1 +√7;

3 7−√7

Soluci´on:

1. √3

7 = 3√7

7

5

√

5 = 5√5

5 =

√

5

2. 1 1 +√7 =

1−√7

(1 +√7)(1−√7) =

1−√7 (1−(√7)2₎ =

1−√7 1−7 =−

1−√7 6

3 7−√7 =

3(7 +√7) (7−√7)(7 +√7) =

3(7 +√7) (72₋₍√₇₎2₎ =

3(7 +√7)

42 =

7 +√7 14

Problema 37 Simplifica todo lo que puedas

√

27−√3 +√192−2√12,

4

√

a3√_a

3

√

a2 ,

√

27 + 1 2

√

12−2√75

Soluci´on:

√

27−√3 +√192−2√12 = 7√3,

4

√

a3√_a

3

√

a2 =

12√

a7_,

√

27 +1 2

√

12−2√75 =−6√3

3 1 +√7;

3 3 √ 3, √ 2 √

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Soluci´on:

3

1 +√7 =−

1−√7 2 ; 3 3 √ 3 = 3 √ 9, √ 2 √

3−√2 =

√

6 + 2

√

27 +1 2

√

12−2

√

75,

√

75√325

√

15 ,

√

48 + 3

√

75−√27 +

√

108

Soluci´on:

√

27 +1 2

√

12−2

√

75 =−6

√

3,

√

75√325

√

15 = 5 6

√

5,

√

48 + 3√75−√27 +√108 = 22√3

2 1 +√5;

2 3

√

32,

√

5

√

2−√3

Soluci´on:

2

1 +√5 =−

1−√5 2 ;

2 3

√

32 =

2√3 3 3 ,

√

5

√

2−√3 =−

√

10−√15

3√32− 1

3

√

72 +√128,

√

27√3 9 6

√

3 ,

√

48 + 3√75−√27 +√108

Soluci´on:

3√32−1

3

√

72 +√128 = 18√2,

√

27√39 6

√

3 = 9,

√

48 + 3√75−√27 +√108 = 22√3

7 2 +√11;

6 5

√

32,

√

7

√

7−√3

Soluci´on:

7

2 +√11 =−2 +

√

11; √₅6

32 = 2

5

√

33_,

√

7

√

7−√3 =

(22)

√

75 +1 2

√

192 +√147,

√

216√39 6

√

3 ,

√

96−√150 + 2√294

Soluci´on:

√

75 +1 2

√

192 +√147 = 16√3,

√

216√3 9 6

√

3 = 18

√

2,

√

96−√150 + 2√294 = 13√6

4 1 +√5;

3 7

√

32,

√

3

√

3−√7

Soluci´on:

4

1 +√5 =−1 +

√

5; √₇3

32 =

7

√

35_,

√

3

√

3−√7 =−

3 +√21 4

1.2.2 Ecuaciones Racionales Problema 45

√

x−1−√x= 4

Soluci´on:

√

x−1 = 4 +√x=⇒x−1 = 16 +x+ 8√x=⇒ −17 = 8√x=⇒x= 289 64

Problema 46

2 +√x−1 =x

Soluci´on:

√

x−1 =x−2 =⇒x−1 =x2+ 4−4x=⇒x2−5x+ 1 = 0 =⇒

    

x= 4,791287847

x= 0,2087121525 No Vale

Problema 47

√

x−1 +√x+ 1 = 3

Soluci´on:

√

(23)

Problema 48

3−√x+ 2 =x

Soluci´on:

−√x+ 2 =x−3 =⇒x+ 2 =x2+ 9−6x=⇒x2−7x+ 7 = 0 =⇒

    

x= 5,79129

x= 1,20871 No Vale

Problema 49

√

x−3 +√x= 4

Soluci´on:

√

x−3 = 4−√x=⇒x−3 = 16 +x−8√x=⇒ −19 =−8√x=⇒x= 361 64

Problema 50

√

x+ 4 =x−1

Soluci´on:

x+ 4 =x2+ 1−2x=⇒x2−3x−3 = 0 =⇒

    

x= 3,7912

x=−0,79128 No Vale

Problema 51

√

2x−1 +x= 8

Soluci´on:

2x−1 = 64 +x2−16x=⇒x2−18x+ 65 = 0 =⇒

(

x= 5

x= 13 No Vale

Problema 52

√

x+ 1 =x−1

Soluci´on:

x+ 1 = 1 +x2−2x=⇒x2−3x= 0 =⇒x(x−3) = 0

(

x= 3

(24)

Problema 53

√

2x+ 3−√x−2 = 2

Soluci´on:

√

2x+ 3 = 2+√x−2 =⇒2x+3 = 4+x−2+4√x−2 =⇒x+1 =−4√x−2

x2+ 2x+ 1 = 16x+ 32 =⇒x2−14x+ 33 = 0

(

x= 3

x= 11

Problema 54

√

3x−5 +x= 1

Soluci´on:

3x−5 = 1 +x2−2x=⇒x2−5x+ 6 = 0

(

x= 3 No Vale

x= 2 No Vale

Problema 55

p

x2₋_{8 =}_x_{+ 2}

Soluci´on:

x2−8 =x2+ 4x+ 4 =⇒x=−3

Problema 56 Halla las soluciones reales de:

√

x+ 6 +√2−x= 4

Soluci´on:

(√x+ 6)2= (4−√2−x)2

x+ 6 = 16 + (√2−x)2−8√2−x

2x−12 =−8√2−x

x−6 =−4√2−x

(x−6)2 = (−4√2−x)2

x2+ 36−12x= 16(2−x)

x2+ 4x+ 4 = 0 =⇒x= −4±

√

16−16

(25)

Problema 57 Halla las soluciones reales de:

√

x−1 +√x= 2

Soluci´on:

(√x−1)2 = (2−√x)2

x−1 = 4 + (√x)2−4√x

x−x−1−4 =−4√x

−5 =−4√x

(−5)2 = (−4√x)2

25 = 16x=⇒x= 25 16

Problema 58 Hallar las soluciones reales de:

√

x+ 7 +√x= 7

Soluci´on:

√

x+ 7 +√x= 7 =⇒√x+ 7 = 7−√x=⇒(√x+ 7)2 = (7−√x)2=⇒

x+ 7 = 49 +x−14√x=⇒ −42 =−14√x=⇒3 =√x=⇒x= 9

√

x+ 6 +√x= 3

Soluci´on:

√

x+ 6 +√x= 3 =⇒√x+ 6 = 3−√x=⇒(√x+ 6)2 = (3−√x)2=⇒

x+ 6 = 9 +x−6√x=⇒ −3 =−6√x=⇒ 1

2 =

√

x=⇒x= 1 4

√

x+ 1−√x−1 = 1

Soluci´on:

√

x+ 1−√x−1 = 1 =⇒√x+ 1 = 1 +√x−1 =⇒

(26)

Problema 61 Calcular:

√

x+ 1−√x−1 = 1

Soluci´on:

√

x+ 1−√x−1 = 1 =⇒x= 5 4

1.3 Logaritmos

1.3.1 Ecuaciones Logar´ıtmicas

Problema 62 Resolver las ecuaciones:

1. logx+ log 50 = log 1000

2. 2 logx3= log 8 + 3 logx

Soluci´on:

1.

logx+ log 50 = log 1000

log(50x) = log 1000

50x= 1000

x= 1000 50 = 20

2.

2 logx3= log 8 + 3 logx

6 logx= log 8 + 3 logx

6 logx−3 logx= log 8

3 logx= log 8

logx3= log 23

x3= 23

x= 2

(27)

1.3. LOGARITMOS 27

1. 3 logx+ 2 logx2 = log 128

2. 3 logx2 = 4 + 4 logx

Soluci´on:

1.

3 logx+ 2 logx2= log 128

3 logx+ 4 logx= log 128

7 logx= log 128

logx7 = log 27

x7 = 27

x= 2

2.

3 logx2= 4 + 4 logx

6 logx−4 logx= 4

2 logx= 4

logx= 2

logx= log 102

x= 102 = 100

Problema 64 Halla las soluciones de:

log(3x2−2) = 1 + log(x−1)

Soluci´on:

log(3x2−2) = log 10 + log(x−1)

log(3x2−2) = log 10(x−1)

3x2−2 = 10(x−1)

3x2−10x+ 8 = 0 =⇒x= 10±

√

100−96 6

x= 10±2

(28)

log(x2+ 6x+ 7) = 1 + log(x+ 1)

Soluci´on:

log(x2+ 6x+ 7) = log 10 +log(x+ 1)

log(x2+ 6x+ 7) = log 10(x+ 1)

x2+ 6x+ 7 = 10(x+ 1)

x2−4x−3 = 0 =⇒x= 3, x= 1

log(3x2−2) = 1 + log(x−1)

Soluci´on:

log(3x2−2) = 1 + log(x−1) =⇒log(3x2−2) = log 10 + log(x−1) =⇒

log(3x2−2) = log 10(x−1) =⇒(3x2−2) = 10(x−1) =⇒3x2−10x+ 8 = 0

=⇒

(

x= 2

x= 4₃

log(x2+ 2699) = 2 + log(x+ 2)

Soluci´on:

log(x2+ 2699) = 2 + lg(x+ 2) =⇒log(x2+ 2699) = log 100 + log(x+ 2) =⇒

log(x2+ 2699) = log 100(x+ 2) =⇒(x2+ 2699) = 100(x+ 2) =⇒

x2−100x+ 2499 = 0 =⇒

(

x= 51

x= 49

Problema 68 Calcular

log(x2−1) + 2 = 1 + 2 log(x+ 1)

Soluci´on:

log(x2−1) + 2 = 1 + 2 log(x+ 1) =⇒log(x2−1) + 1 = 2 lg(x+ 1) =⇒

lg 10(x2−1) = lg(x+ 1)2 =⇒10(x2−1) = (x+ 1)2 =⇒9x2−2x−11 = 0

=⇒

  

x=−1

(29)

1.3. LOGARITMOS 29

Problema 69 Resolver la siguiente ecuaci´on:

log(1 +x2)−1 = log(x−2)

Soluci´on:

log(1 +x2)−1 = log(x−2) =⇒log(1 +x2)−log 10 = log(x−2) =⇒

log 1 +x

2

10

!

= log(x−2)

1 +x2

10 =x−2 =⇒1 +x

2 _{= 10}_x_{+ 10 =}_⇒_x2₋₁₀_x_{+ 21 = 0 =}_⇒

x= 7, x= 3

1. log10

x = 2−2 logx

2. 3 logx−2 = 2 logx

Soluci´on:

1.

log 10−logx= 2−2 logx

1−logx= 2−2 logx

2 logx−logx= 2−1

logx= 1 =⇒x= 10

2.

3 logx−2 = 2 logx

3 logx−2 logx= 2

logx= 2 =⇒x= 102= 100

1. log 10(x+ 2)−log(x2) = 1

2. logx+ logx2 = 3

(30)

1.

log10(x+ 2)

x2 = log 10

10(x+ 2)

x2 = 10

10x+ 20 = 10x2

x2−x−2 = 0 =⇒x= 2, x=−1

2.

logx+ 2 logx= 3

3 logx= 3

logx= 1 =⇒x= 10

Problema 72

log(3x+ 1)−2 logx= 2

Soluci´on:

log

₃_x_{+ 1}

x2

= log 100 =⇒100x2−3x−1 = 0 =⇒

(

x= 0,116187

x=−0,0861187 No Vale

Problema 73

log(2x+ 1)−2 logx= 1

Soluci´on:

log

₂_x_{+ 1}

x2

= log 10 =⇒10x2−2x−1 = 0 =⇒

(

x= 0,43166

x=−0,23166 No Vale

Problema 74

2 log(x+ 1)−logx= 1

Soluci´on:

log (x+ 1)

2

x

!

= log 10 =⇒x2−8x+ 1 = 0 =⇒

(

x= 0,127

(31)

1.3. LOGARITMOS 31

Problema 75

logx−log(1−x) = 2

Soluci´on:

log

_x

1−x

= log 100 =⇒101x= 100 =⇒x= 100 101

Problema 76

log(x+ 1)−log(x2−1) = 1

Soluci´on:

log

_x_{+ 1}

x2₋₁

= log 10 =⇒10x2−x−11 = 0 =⇒

(

x= 1,1

x=−1 No Vale

Problema 77

logx−log(1−x) = 2

Soluci´on:

log

_x

1−x

= log 100 =⇒101x= 100 =⇒x= 100 101

1. logx2−log(x−1) + 1 = 2 logx

2. log(x+ 1)−2 log(x−1) = 1

Soluci´on:

1. logx2−log(x−1) + 1 = 2 logx=⇒log 10x

2

x−1 = logx

2 ₌_⇒

x2(11−x) = 0 =⇒x= 11 yx= 0 (no vale).

2. log(x+ 1)−2 log(x−1) = 1 =⇒log x+ 1

(x−1)2 = log 10 =⇒

10x2−21x+ 9 = 0 =⇒x= 3 5 yx=

3

5 no vale

1. log(10x2−2)−1 = log(x+ 1) + logx

2. log(3x2₋₂₎₋_{2 log(1}₋_x_{) = 1}

(32)

1. log(10x2−2)−1 = log(x+ 1) + logx=⇒log10x

2₋₂

10 = logx(x+ 1)

=⇒10x2−2 = 10x(x+ 1) =⇒x=−1

5

2. log(3x2−2)−2 log(1−x) = 1 =⇒log 3x

2₋₂

(1−x)2 = log 10 =⇒

7x2₋₂₀_x_{+ 12 = 0 =}_⇒_x₌ 6

7, x= 2 (no vale)

1. 2 log(x−1) + 1 = log(x2−1)

2. log(10(x3+ 2x))−2 log(x+ 1) = 1 + logx

Soluci´on:

1. 2 log(x−1) + 1 = log(x2−1) =⇒log 10(x−1)2 = log(x2−1)

=⇒9x2−20x+ 11 = 0 =⇒x= 11

9 yx= 1 (no vale).

2. log(10(x3+ 2x))−2 log(x+ 1) = 1 + logx=⇒

log10(x

3_{+ 2}_x₎

(x+ 1)2 = log 10x=⇒2x

2₋_x_{= 0 =}_⇒

x= 1

2 yx= 0 (no vale).

1. log(x−1) + log(x+ 1) = 2 logx−1

2. logx2+ 3 logx= 2

Soluci´on:

1. log(x−1) + log(x+ 1) = 2 logx−1 =⇒log(x2−1) = log(x−1)2

=⇒9x2 = 10 =⇒x=

√

10

3 , x=−

√

10

3 (no vale)

2. logx2+ 3 logx= 2 =⇒logx5= log 100 =⇒x= √5100 = 2,51188

Problema 82 Resolver la siguiente ecuaci´on:

log(2 +x)−logx= 1 + log(1−x)

Soluci´on:

log(2 +x)−logx= 1 + log(1−x) =⇒log2 +x

(33)

1.3. LOGARITMOS 33

log2 +x

x = log(10(1−x)) =⇒10x

2₋₉_x_{+ 2 = 0 =}_⇒ _x₌ 1

2, x= 2 5

Problema 83 Unos problemas para ejercitarse:

1. 5 log 2x= 20 Sol: x= 5000

2. 3 log 5x=−9 Sol: x= 0,0002

3. log2x₅−4 = 2 Sol: x= 252

4. log(x+ 1)2 = 2 Sol: x= 9; x=−11

5. log(7x+ 15)−log 5 = 1 Sol: x= 5

6. logx₂ = 1 + log(21−x) Sol: x= 20

7. log10_x = 2−2 logx Sol: x= 10; x= 0

8. 2 logx−log(x2−2x+ 6) = 0 Sol: x= 3

9. log(2x−3) + log(3x−2) = 2−log 25 Sol: x= 2; x= 1₆

10. log(3x2−2) = 1 + log(x−1) Sol: x= 2; x= 4₃

11. logx2+ 3 logx= 2 Sol: x= 1025

12. 2 logx2−2 logx= 2 Sol: x= 10

13. logx2+ 1 = logx3 Sol: x= 10

14. log(1−x) + logx= 1 Sol: No tiene soluci´on real.

15. logx−log(1−x) = 1 Sol: x= 10₁₁

16. logx+ 2 = logx3 Sol: x= 10

17. log(1 +x) + log(1−x) = 2 Sol: No tiene soluci´on real.

18. log(2x+ 7)−log(x−1) = log 5 Sol: x= 4

19. log(35_log(5−₋_xx2₎) = 3 Sol: x= 3 : x= 2 20. logx2₋_log10x+11

10 = 1 Sol: x= 11; x=−1

21. log(2x+ 2) + log(x+ 3) = log 6 Sol: x= 0, x=−4

22. log 2+log(_log(2_x₋x₂₎2−2) = 2 Sol: x= 2

23. log(x+ 6)−1₂log(2x−3) = 2−log 25 Sol: x= 6; x= 14

(34)

25. 2 logx= 2 + logx Sol: x= 0; x= 2

26. log 8 + (x2−5x+ 7) log 3 = log 24 Sol: x= 3; x= 2

27. 2 logx−log 16 = logx₂ Sol: x= 0; x= 8

28. log(2x+4)+log(3x+1)−log 4 = 2 log(8−x) Sol: x=−42 x= 3

29. log(35_log(5−₋x_x3₎) = 3 Sol: x= 3 x= 2 30. log 2+log(11_log(5₋_x₎−x2) = 2 Sol: x= 1₃ x= 3

31. log(5x+ 4)−log 2 = 1₂log(x+ 4) Sol: x= 0

32. (x2−x+ 3) log 4 = 3 log1₄ Sol: No tiene soluci´on.

1.3.2 Sistemas de Ecuaciones Logar´ıtmicas

Problema 84 Resolver el sistema de ecuaciones logar´ıtmicas:

(

3 logx+ 2 logy= 12 logx_y = −1

Soluci´on:

(

3 logx+ 2 logy= 12 logx_y = −1 =⇒

(

3 logx+ 2 logy = 12 logx− logy = −1

Haciendo el cambio de variables logx=uy logy=v el sistema quedar´a de la siguiente forma:

(

3u+ 2v= 12

u− v= −1 =⇒

(

3u+ 2v = 12 2u− 2v = −2 =⇒

(

u= 2

v= 3

Deshaciendo el cambio de variables nos quedar´ıa:

(

logx=u= 2 logy =v= 3 =⇒

(

logx= log 102 logy= log 103 =⇒

(

x= 100

y= 1000

(

2 logx+ logy= 4 logx_y = −1

Soluci´on:

(

2 logx+ logy= 4 logx_y = −1 =⇒

(

(35)

1.3. LOGARITMOS 35

Haciendo el cambio de variables logx=uy logy=vel sistema quedar´a de la siguiente forma:

(

2u+ v= 4

u− v= −1 =⇒

(

u= 1

v= 2

(

logx=u= 1 logy=v= 2 =⇒

(

logx= log 101 logy = log 102 =⇒

(

x= 10

y= 100

(

logx_y32 = 1 log(x2y) = 2

Soluci´on:

(

logx_y32 = 1 log(x2y) = 2 =⇒

(

3 logx− 2 logy= 1 2 logx+ logy= 2

(

3u− 2v= 1 2u+ v= 2 =⇒

(

3u− 2v= 1 4u+ 2v= 4 =⇒

(

u= 5₇

v= 4₇

(

logx=u= 5₇ logy=v= 4₇ =⇒

(

x= 1057

y= 1047 =⇒

(

x= 5,179474679

y= 3,72759372

(

logx_y4 = 1 log(x·y2) = 2

Soluci´on:

(

logx_y4 = 1 log(x·y2) = 2 =⇒

(

4 logx− logy = 1 logx+ 2 logy = 2

(

4u− v= 1

u+ 2v= 2 =⇒

(

4u− v= 1

−4u− 8v= −8 =⇒

(

u= 4₉

v= 7₉

(

logx=u= 4₉ logy=v= 7₉ =⇒

(

x= 1049

y= 1079 =⇒

(

x= 2,782559402

(36)

(

logx_y23 = 1 log(x2y) = 2

Soluci´on:

(

logx_y32 = 1 log(x2y) = 2 =⇒

(

3 logx− 2 logy = 1 2 logx+ logy = 2

(

3u− 2v= 1 2u+ v= 2 =⇒

(

3u− 2v= 6 4u+ 7v= 4 =⇒

(

u= 9₇

v= 4₇

(

logx=u= 5₇ logy =v= 4₃ =⇒

(

x= 1053

y= 1047 =⇒

(

x= 5,179474679

y= 2,71755372

(

logx_y23 = 2 log(x2_y₎ ₌ ₃

Soluci´on:

(

logx_y32 = 2 log(x2y) = 3 =⇒

(

3 logx− 2 logy = 2 2 logx+ logy = 3

(

3u− 2v= 2 2u+ v= 3 =⇒

(

3u− 2v= 2 4u+ 2v= 6 =⇒

(

u= 8₇

v= 5₇

(

logx=u= 8₇ logy =v= 5₇ =⇒

(

x= 1087

y= 1057 =⇒

(

x= 13,89495494

y= 5,179474679

  

log(x·y) = 3 logx

(37)

1.3. LOGARITMOS 37

Soluci´on:

  

log(x·y) = 3 logx

y = 1

=⇒

(

logx+ logy= 3 logx− logy= 1

(

u+ v= 3

u− v= 1 =⇒

(

u= 2

v= 1

(

logx=u= 2 logy=v = 1 =⇒

(

x= 102 = 100

y= 10

(

logx−logy2 = 3 log(x2·y) = 1

Soluci´on:

(

logx−logy2 = 3 log(x2·y) = 1 =⇒

(

logx− 2 logy = 3 2 logx+ logy = 1

(

u− 2v = 3 2u+ v = 1 =⇒

(

u= 1

v=−1

(

logx=u= 1

logy =v=−1 =⇒

(

x= 101= 10

y= 10−1 = 0,1 Problema 92 Resolver el sistema de ecuaciones logar´ıtmicas:

(

log_yx2 = 1 log(x2y) = 2

Soluci´on:

(

log_yx2 = 1 log(x2y) = 2 =⇒

(

(38)

(

u− 2v= 1 2u+ v= 2 =⇒

(

u= 1

v= 0

(

logx=u= 1 logy=v= 0 =⇒

(

x= 10

y= 1

(

log_yx2 = 3 log(x2y) = 2

Soluci´on:

(

log_yx2 = 3 log(x2y) = 2 =⇒

(

logx− 2 logy = 3 2 logx+ logy = 2

(

u− 2v= 3 2u+ v= 2 =⇒

(

u= 7₅

v =−4₅

(

logx=u= 7₅

logy=v =−4₅ =⇒

(

x= 107/5 _{= 25}_,_11886431

y= 10−4/5 _{= 0}_,_1584893192

Problema 94

    

log(xy2) = 2

log x

2

y

!

= 3

Soluci´on:

(

logx+ logy = 2 2 logx− logy = 3 =⇒

(

u+ v= 2 2u− v= 3 =⇒

(

u= logx= 8/5 =⇒x= 39,81

(39)

1.3. LOGARITMOS 39

Problema 95

  

log(x2y) = 3

log

_x

y

= 2

Soluci´on:

(

2 logx+ logy= 3 logx− logy= 2 =⇒

(

2u+ v= 3

u− v= 2 =⇒

(

u= logx= 5/3 =⇒x= 46,41589

v= logy=−1/3 =⇒y= 0,464159

Problema 96

  

2 log(xy) = 3

log

_x

y2

= 5

Soluci´on:

(

2 logx+ 2 logy= 3 logx− 2 logy= 5 =⇒

(

2u+ 2v= 3

u− 2v= 5 =⇒

(

u= logx= 8/3 =⇒x= 464,1588

v= logy=−7/6 =⇒y= 0,068129

Problema 97

(

logx+ logy= 3 2 logx− logy= 0

Soluci´on:

(

logx+ logy= 3 2 logx− logy= 0 =⇒

(

u+ v= 3 2u− v= 0 =⇒

(

u= logx= 1 =⇒x= 10

v= logy= 2 =⇒y= 100

Problema 98

(

log(x3y2) = 8 logx_y = 1 Soluci´on:

(

3 logx+ 2 logy= 8 logx− logy= 1 =⇒

(

3u+ 2v = 8

u− v = 1 =⇒

(

u= logx= 2 =⇒x= 100

(40)

Problema 99

(

logx+ 2 logy= 3

−logx+ logy= 0

Soluci´on:

(

logx+ 2 logy= 3

−logx+ logy= 0 =⇒

(

u+ 2v= 3

−u+ v= 0 =⇒

(

u= logx= 1 =⇒x= 10

v= logy= 1 =⇒y= 10

        

log(xy)2 _{= 4}

log x 3 y2 ! = 1 Soluci´on:         

log(xy)2 = 4

log x 3 y2 ! = 1 =⇒ (

2 logx+ 2 logy = 4 3 logx−2 logy = 1 =⇒

(

2u+ 2v = 4 3u−2v = 1

=⇒

(

u= logx= 1

v= logy= 1 =⇒

(

x= 10

y= 10

      

log(xy)2 = 4

log _x y2 = 2 Soluci´on:       

log(xy)2 = 4

log _x y2 = 2 =⇒ (

2 logx+ 2 logy = 4 logx−2 logy = 2 =⇒

(

2u+ 2v = 4

u−2v = 2

=⇒

(

u= 2 = logx v= 0 = logy =⇒

(

x= 100

(41)

1.3. LOGARITMOS 41

      

log(xy)2 = 8

log _x y2 = 4 Soluci´on:       

log(xy)2 = 8

log _x y2 = 4 =⇒     

2 logx+ 2 logy= 8

logx− 2 logy= 4

=⇒     

2u+ 2v= 8

u− 2v= 4 =⇒     

u= 4

v= 0 =⇒     

u= 4 = logx=⇒x= 10000

v= 0 = logy =⇒y = 1

1.

(

2 logx− 5 logy= −1 3 logx+ 2 logy= 8

Sol: x= 100; y= 10 2.

(

4 logx−3 logy = −1 log(x·y) = 5

Sol: x= 100; y= 1000 3.

(

logx+ logy3 = 5 logx_y32 = 4

Sol: x= 100; y= 10 4.

(

log(x2·y) = 2 logx_y = 1

Sol: x= 10; y= 1 5.

(

logx2−3 logy =−1 log(x·y2) = 3

(42)

6.

(

logx2−3 logy = 2 log(_yx2) = 3

Sol: x= 10−5; y= 10−4 7.

(

logx−logy = 7 logx+ logy = 3

Sol: x= 105; y= 10−2 8.

(

x−y = 15 logx+ logy = 2

Sol: x=−5; y =−20 o bienx= 20; y= 5

9.

(

logx+ 3 logy = 5 logx_y2 = 3

Sol: x= 100; y= 10 10.

(

2 logx2 −logy2 = 4 2 logx + logy2 = 2

Sol: x= 100; y= 1

1.4 Exponenciales

1.4.1 Ecuaciones Exponenciales:

3x2+5x−4·92x+3= 27x−1 Soluci´on:

3x2+5x−4·32(2x+3)= 33(x−1) 3x2+5x−4+2(2x+3) = 33(x−1)

x2+ 5x−4 + 4x+ 6 = 3x−3

x2+ 6x+ 5 = 0 =⇒x= −6±

√

36−20 4

x= −6±4

(43)

1.4. EXPONENCIALES 43

3x2+5x−4·92x+3= 27x−1 Soluci´on:

3x2+5x−4·32(2x+3) = 33(x−1) 3x2+5x−4+2(2x+3)= 33(x−1)

x2+ 5x−4 + 4x+ 6 = 3x−3

x2+ 6x+ 5 = 0 =⇒x= −6±

√

36−20 4

x= −6±4

2 =⇒x=−1, x=−5

Problema 106 Calcular

2·32x−1+ 3x+1−1 = 0 Soluci´on:

2·32x−1+ 3x+1−1 = 0 =⇒ 2·3

2x

3 + 3·3

x₋_{1 = 0 =}_⇒₂_·₃2x_{+ 9}_·₃x₋_{3 = 0}

Haciendo el cambio de variables u= 3x la ecuaci´on quedar´a de la siguiente forma:

2u2+ 9u−3 = 0 =⇒u= 0,3117376914, u=−4,811737691

Deshaciendo el cambio de variable tenemos que

u= 0,3117376914 = 3x=⇒log 0,3117376914 = log 3x=⇒

xlog 3 = log 0,3117376914 =⇒

x= log 0,3117376914

log 3 =−1,060968632

En el otro caso, u=−4,811737691 = 3x no es posible obtener soluci´on. Problema 107

72x−1+ 7x+1−1 = 0 Soluci´on:

(7x)2

7 + 7·7

x₋_{1 = 0 =}_⇒ t2

7 + 7t−1 = 0 =⇒

(

t= 0,14244

t=−49,14224

(

t= 0,14244 = 7x =⇒x=−1,0015

(44)

Problema 108

62x−1+ 6x+1−1 = 0 Soluci´on:

(6x)2 6 + 6·6

x₋_{1 = 0 =}_⇒ t2

6 + 6t−1 = 0 =⇒

(

t= 0,027764

t=−36,02776

(

t= 0,027764 = 6x =⇒x=−2,0004

t=−36,02776 = 6x=⇒ No Vale

Problema 109

32x+1−3x−1−1 = 0 Soluci´on:

3(3x)2− 3

x

3 −1 = 0 =⇒3t

2₋ t

3 −1 = 0 =⇒

(

t= 0,63557

t=−0,524461

(

t= 0,63557 = 3x=⇒x=−0,41255

t=−0,524461 = 3x₌_⇒ _{No Vale}

Problema 110

2x−2x+1+ 1 = 0

Soluci´on:

2x−2·2x+ 1 = 0 =⇒t−2t+ 1 = 0 =⇒ t= 1

t= 2x= 1 =⇒ x= 0

Problema 111

52x−1−5x+ 1 = 0 Soluci´on:

(5x₎2

5 −5

x_{+ 1 = 0 =}_⇒ t2

5 −t+ 1 = 0 =⇒ t

2₋₅_t_{+ 5 = 0}

(

t= 5x= 3,618 =⇒x= 0,714

(45)

Problema 112

2x−2x−1−1 = 0 Soluci´on:

2x−2

x

2 −1 = 0 =⇒t−

t

2 −1 = 0 =⇒ t= 2 =⇒2

x_{= 2 =}_⇒_x_{= 1}

1. 2x+1 = 8 Sol: x= 2

2. 2x+3+ 4x+1 = 320 Sol: x= 3

3. 612−3x= 216 Sol: x= 3 4. 53x−12_{= 125} _Sol: _x_{= 5}

5. 2x_{+ 2}x+3 _{= 36} _Sol: _x_{= 2}

6. 3x+ 3x−2 = 270 Sol: x= 5

7. 5x+ 5x+1+ 5x+2= 31₂₅ Sol: x=−2

8. 52x2+3x−11= 125 Sol: x= 2; x=−7 2

9. 4x+ 22x−1 = 24 Sol: x= 2; la otra soluci´on no es real. 10. 2x+ 22x = 6 Sol: x= 1; la otra soluci´on no es real.

11. 3x+3+ 9x+2 = 4 Sol: x=−2; la otra soluci´on no es real.

12. 42x+1−4x+2= 768 Sol: x= 2; la otra soluci´on no es real.

13. 2x·3x= 12·18 Sol: x= 3

14. 9x+3 _{= 3}2x+5 _{Sol: No tiene soluci´}_on.

15. 8x2+3x+2_{= 1} _Sol: _x₌₋_1; _x₌₋₂

16. 5x+ 5x−1+xx−2= 31 Sol: x= 2 17. 2x+2 = 0,52x−1 Sol: x=−1

3

18. 3

√

a7−x ₌_a2 _Sol: _x_{= 1}

19. 4x−5·2x+ 4 = 0 Sol: x= 2; x= 0

20. 72x+3−8·7x+1+ 1 = 0 Sol: x=−1; x=−2

(46)

22. 10x2−11x+30= (2·5)2 Sol: x= 7; x= 4 23. 3x−1+ 3x+ 3x+1 = 117 Sol: x= 3

24. 32(x+1)−28·3x+ 3 = 0 Sol: x=−2; x= 1

25. 22x−3·2x+1+ 8 = 0 Sol: x= 2; x= 1

26. 2₇5 = 3,5x+1 Sol: x=−6 27. 5x− 5

5x−1 −24 = 0 Sol: x= 2

28. 43−x2−x

= 1 Sol: x= 3; x= 2

29. 21−x2 = 1₈ Sol: x=±2 30. 32x−1 = 3

q

9x2−14 Sol: x= 11

2 ; x=

1 2

31. 3·2x+3= 192·3x−3 Sol: No tiene soluci´on. Problema 114 M´as problemitas:

1. 2x−2+ 2x+1−1 = 0 Sol: x=−1,169925001 2. 3x+1+ 3x−3x−1 = 2 Sol: x=−0,5517286062 3. 2x−2−2x+ 2x−1 = 0 Sol: No tiene soluci´on. 4. 3x−2+ 2·3x= 1 Sol: x=−0,6801438331 5. 4x−1−3·4x+ 4x−2 = 0 Sol: No tiene soluci´on. 6. 22x−1+ 2x+1−2 = 0 Sol: x=−0,2715533031 7. 52x−1_{+ 3}_·₅x₋_{2 = 0} _Sol: _x₌₋₀_,_2778665354

8. 32x−2+ 3x−1−1 = 0 Sol: x=−1,011034949 9. 22x+1−3·2x−1−3 = 0 Sol: x= 0,7275884076 10. 22x−1−3·2x+2−2 = 0 Sol: x= 4,594878436 11. 72x−1₋₇x+1₋_{2 = 0} _Sol: _x_{= 2}_,_002970617

12. 62x−1−6x−1−4 = 0 Sol: x= 0,9437163029 13. 54x−1−52x+1−3 = 0 Sol: x= 0,4606479652 14. 44x−1−42x+1−7 = 0 Sol: 1,034204992

(47)

16. 34x+1+ 2·32x−2−2 = 0 Sol: x=−0,1129051332 17. 24x+2+ 3·22x−1 = 0 Sol: x=−1

18. 54x−2+ 52x−1 = 0 Sol: x=−0,01174112826 1.4.2 Sistemas de Ecuaciones Exponenciales:

Problema 115 Resolver el sistema de ecuaciones exponenciales:

(

4x+1− 6y = 40 2·4x− 6y = −88

Soluci´on

(

4x+1− 6y = 40 2·4x− 6y = −88 =⇒

(

4·4x− 6y = 40 2·4x− 6y = −88

Haciendo el cambio de variables 4x = u y 6y = v el sistema quedar´a de la siguiente forma:

(

4u− v= 40 2u− v= −88 =⇒

(

u= 64

v= 216

(

4x=u= 64 6y =v= 216 =⇒

(

4x = 43 6y = 63 =⇒

(

x= 3

y= 3

(

2x+ 5y = 9 2x+2+ 5y+1= 41

Soluci´on

(

2x+ 5y = 9 2x+2+ 5y+1= 41 =⇒

(

2x+ 5y = 9 4·2x+ 5·5y = 41

(

u+ v= 9 4u+ 5v= 41 =⇒

(

u= 4

v= 5

(

2x=u= 4 5y =v= 5 =⇒

(

2x= 22 5y = 51 =⇒

(

x= 2

(48)

(

2x−1+ 3y+1 = 4 2x+1− 3y+1 = 5

Soluci´on:

(

2x−1+ 3y+1 = 4 2x+1− 3y+1 = 5 =⇒

( ₂x

2+ 3·3

y ₌ ₄

2·2x− 3·3y = 5

Haciendo el cambio de variables 2x ₌ _u _{y 3}y ₌ _v _{el sistema quedar´}_{a de la}

siguiente forma:

( _u

2+ 3v= 4

2u− 3v= 5 =⇒

(

u= 18₅

v= 11₁₅

(

2x =u= 18₅ 3y =v= 11₁₅ =⇒

(

xlog 2 = log18₅

ylog 3 = log11₁₅ =⇒

  

x= log 18

5

log 2 = 1,847996906

y= log 11 15

log 3 =−0,2823151820

(

2x+1− 3y−1 = 4 2x+1₊ ₃y+1 ₌ ₅

Soluci´on:

(

2x+1− 3y−1 = 4 2x+1+ 3y+1 = 5 =⇒

(

2·2x− 3y

3 = 4

2·2x₊ ₃_·₃y ₌ ₅

(

2u− v

3 = 4

2u+ 3v= 5 =⇒

(

u= 41₂₀ = 2,05

v = ₁₀3 = 0,3

(

2x =u= 41₂₀ 3y =v= ₁₀3 =⇒

(

xlog 2 = log41₂₀

ylog 3 = log₁₀3 =⇒

  

x= log 41 20

log 2 = 1,035623909

y= log 3 10

log 3 =−1,095903274

(

(49)

Soluci´on:

(

2x+1− 3y−1= 3 2x+1+ 3y+1= 4 =⇒

(

2·2x− 3y

3 = 3

2·2x+ 3·3y = 4

(

2u− v₃ = 3 2u+ 3v= 4 =⇒

(

u= 31₂₀ = 1,55

v= ₁₀3 = 0,3

(

2x=u= 31₂₀ 3y =v= ₁₀3 =⇒

(

xlog 2 = log31₂₀

ylog 3 = log₁₀3 =⇒

  

x= log 31 20

log 2 = 0,6322682154

y= log 3 10

log 3 =−1,095903274

(

2x+1− 3y−1= 1 2x−1₊ ₃y+1₌ ₂

Soluci´on:

(

2x+1− 3y−1 = 1 2x−1+ 3y+1 = 2 =⇒

        

2·2x+ 3

y

3 = 1

2x

2 − 3·3

y ₌ ₂

Haciendo el cambio de variables 2x = u y 3y = v el sistema quedar´a de la siguiente forma:       

2u− v

3 = 1

u

2+ 3v= 2 =⇒

(

6u− v= 3

u+ 6v= 4

(

u= 22₃₇

v= 21₃₇

(

2x=u= 22₃₇ 3y =v= 21₃₇ =⇒

(

xlog 2 = log22₃₇

ylog 3 = log21₃₇ =⇒

  

x= log 22 37

log 2 =−0,7500217469

y= log 21 37

log 3 =−0,5155553790

Problema 121

(

(50)

Soluci´on:        3x

3 + 2·2

y _{= 2}

3x− 2y = 3 =⇒        u

3+ 2v = 2

u− v = 3 =⇒         

u= 27 7 = 3

x ₌_⇒_x_{= 1}_,₂₂₈₇₆

v= 13 7 = 2

y ₌_⇒_y_{= 0}_,₈₉₃₈₅

Problema 122

(

2x−1₊ ₃y+1 _{= 5}

2x− 3y = 2

Soluci´on:        2x

2 + 3·3

y _{= 5}

2x− 3y = 2 =⇒        u

2+ 3v = 5

u− v = 2 =⇒         

u= 22 7 = 2

x ₌_⇒_x_{= 1}_,₆₅₂₀₈

v= 8 7 = 3

y ₌_⇒_y _{= 0}_,₁₂₁₅₄

Problema 123

(

3x−2+ 2y = 1 2x+ 3·2y = 5

Soluci´on:        3x 9 + 2

y _{= 1}

3x+ 3·2y = 5 =⇒        u

9+ v = 1

u+ 3v = 5 =⇒       

u= 3 = 3x=⇒x= 1

v= 2 3 = 2

(51)

Problema 124

(

2x₋ ₃y _{= 1}

2x+ 3y = 3

Soluci´on:

(

2x− 3y = 1 2x+ 3y = 3 =⇒

    

u− v = 1

u+ v = 3 =⇒

    

u= 2 = 2x=⇒x= 1

v= 1 = 3y =⇒y= 0

Problema 125

(

2x+2− 3y = 1 2x+ 2·3y = 3

Soluci´on:

(

2·2x− 3y = 1 2x+ 2·3y = 3 =⇒

    

2u− v = 1

u+ 2v = 3 =⇒

    

u= 1 = 2x=⇒x= 0

v= 1 = 3y =⇒y= 0

Problema 126

(

2x+ 3y = 2 2x+1− 3y = 1

Soluci´on:

(

2x+ 3y = 2 2·2x− 3y = 1 =⇒

    

u+ v = 2

2u− v = 1 =⇒

    

u= 1 = 2x=⇒x= 0

v= 1 = 3y =⇒y= 0

Problema 127 Unos problemas para ejercitarse: 1.

(

3·2x− 4·7y = −172 7·2x₊ ₂_·₇y ₌ ₁₅₄

(52)

2.

(

4x+1− 6y = 40 2·4x− 6y = −88

Sol: x= 3; y= 3 3.

(

2·3x+1− 5y+2 = −2639 4·3x+ 5y = 449

Sol: x= 4; y= 4 4.

(

3x+ 2y = 31 3x+1₋ ₂y+2₌ ₆₅

Sol: x= 3; y= 2 5.

(

5x+y = 253 3x−y = 25 Sol: x= 4; y= 2

6.

(

15· 5x−1 ₋ ₆y ₌ ₃₃₉

3· 5x + 2· 6y+1= 807

Sol: x= 3; y= 2 7.

(

ax+y = a4 ax−y = a2

Sol: x= 3; y= 1

8.

(

8y · 22x = 128 32y · 3x−1 = 27 Sol: x=−70; y= 49

9.

(

33x−y =

√

310

32x+y = 3

(53)

1.5. ECUACIONES LOGAR´ITMICAS Y EXPONENCIALES 53

10.

(

3·2x− 2·3y = −6 4·2x₋ ₃_·₃y ₌ ₋₁₁

Sol: x= 2; y= 2 11.

(

3·2x− 5·3y = 3 2x+1+ 3y+1= 59

Sol: x= 4; y= 2 12.

(

2x− 3y−1= 5 2x+1+ 8·3y = 712

Sol: x= 5; y= 4 13.

(

2·3x₊ ₂y+3 ₌ ₈₆

3x− 2y = 23

Sol: x= 3; y= 2 14.

(

2x+2y = 32 52x−y = 1 Sol: x= 1; y= 2

1.5 Ecuaciones Logar´ıtmicas y Exponenciales

1. log(3x+ 1)−logx= 1 + logx

2. 22x−1+ 2x+2−1 = 0 Soluci´on:

1.

log(3x+ 1)−logx= 1 + logx=⇒log(3x+ 1) = log 10 + 2 logx

log(3x+ 1) = log(10x2) =⇒10x2−3x−1 = 0 =⇒ x= 1

2, x=− 1 5 De las dos soluciones hay una que no es posible ya que no existen logaritmos de n´umeros negativos, es decir, de las dos soluciones la ´

(54)

2.

22x−1+ 2x+2−1 = 0 =⇒ 2

2x

2 + 2

x₂2₋_{1 = 0 =}_⇒₂2x_{+ 8}_·₂x₋_{2 = 0}

Haciendo el cambio de variables u = 2x la ecuaci´on quedar´a de la siguiente forma:

u2+ 8u−2 = 0 =⇒u= 0,2426406871, u=−8,242640687

u= 0,2426406871 = 2x=⇒log 0,2426406871 = log 2x=⇒

xlog 2 = log 0,2426406871 =⇒

x= log 0,2426406871

log 2 =−2,242640687

En el otro caso,u=−8,242640687 = 2x_{no es posible obtener soluci´}_on.

1. log(3x+ 1)−logx= 1 + log(1−x)

2. 22x−1+ 2x+3−1 = 0 Soluci´on:

1.

log(3x+1)−logx= 1+log(1−x) =⇒log(3x+1) = log 10+log(1−x)+logx

log(3x+1) = log(10x(1−x)) =⇒10x2−7x+1 = 0 =⇒ x= 1 2, x=

1 5

2.

22x−1+ 2x+3−1 = 0 =⇒ 2

2x

2 + 2

x_·₂3₋_{1 = 0 =}_⇒₂2x_{+ 16}_·₂x₋_{2 = 0}

u2+ 16u−2 = 0 =⇒u=−16,12403840, u= 0,1240384046

u= 0,1240384046 = 2x=⇒log 0,1240384046 = log 2x=⇒

xlog 2 = log 0,1240384046 =⇒

x= log 0,1240384046

log 2 =−3,011141219

(55)

1. log(3x+ 1)−logx= 1 + log(1−x)

2. 22x−1+ 2x+2−1 = 0 Soluci´on:

1.

log(3x+1)−logx= 1+log(1−x) =⇒log(6x+1) = log 10+log(1−x)+logx

log(3x+1) = log(10x(1−x)) =⇒10x2−8x+1 = 0 =⇒ x= 1 2, x=

1 5

2.

22x−1+3·2x+2−5 = 0 =⇒ 2

2x

2 +3·2

x_·₂2₋_{2 = 0 =}_⇒ 22x

2 +12·2

x₋_{2 = 0}

u2+ 24u−4 = 0 =⇒u= 14,1655, u=−0,331

u= 14,1655 = 2x=⇒log 24,1655 = log 2x =⇒

xlog 2 = log 24,1655 =⇒

x= log 24,1655

log 2 = 4,7948

En el otro caso,u=−0,531 = 2x no es posible obtener soluci´on.

1. log(3x2−2) = 1 + log(x−1)

2. 22x−1+ 2x+1−2 = 0 Soluci´on:

1.

log(3x2−2) = 1 + log(x−1) =⇒log(3x2−2) = log 10 + log(x−1)

(56)

2.

22x−1+ 2x+1−2 = 0 =⇒ 2

2x

2 + 2·2

x₋_{2 = 0 =}_⇒₂2x_{+ 4}_·₂x₋_{4 = 0}

u2+ 4u−4 = 0 =⇒u= 0,8284271247, u=−4,828427124

u= 0,8284271247 = 2x=⇒log 0,8284271247 = log 2x=⇒

xlog 2 = log 0,8284271247 =⇒

x= log 0,8284271247

log 2 =−0,2715533031

En el otro caso,u=−4,828427124 = 2xno es posible obtener soluci´on.

Problema 132 Calcular:

1. log(x2+ 2)−logx= 1

2. 4x−1+ 2x−1 = 0 Soluci´on:

1. log(x2+ 2)−logx= 1 =⇒ x= 0,2041684766, x= 9,795831523

2. 4x−1+ 2x−1 = 0 =⇒ x=−0,2715533031 Problema 133 Resolver las ecuaciones:

1. log(x−1)−log(x+ 1) = 1−logx

2. logx+ 1 = logx2

3. 32x−1+ 3x+1−2 = 0 4. 3x+1+ 3x−1−1 = 0 Soluci´on:

1.

log(x−1)−log(x+ 1) = 1−logx=⇒logx−1

x+ 1 = log 10

x x−1

x+ 1 = 10

x =⇒x

2₋₁₁_x₋_{10 = 0 =}_⇒

x= 11,84428877, x=−0,8442887702

De las dos soluciones hay una que no es posible ya que no existen logaritmos de n´umeros negativos, es decir, de las dos soluciones la ´

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2.

logx+1 = logx2 =⇒log 10x= logx2 =⇒10x=x2=⇒x= 0, x= 10

De las dos soluciones hay una que no es posible ya que no existen logaritmos del cero, es decir, de las dos soluciones la ´unica posible es

x= 10 3.

32x−1+ 3x+1−2 = 0 =⇒ 3

2x

3 + 3·3

x₋_{2 = 0 =}_⇒₃2x_{+ 9}_·₃x₋_{6 = 0}

u2+ 9u−6 = 0 =⇒u= 0,6234753829, u=−9,623475382

u= 0,6234753829 = 3x=⇒log 0,6234753829 = log 3x =⇒

xlog 3 = log 0,6234753829 =⇒

x= log 0,6234753829

log 3 =−0,4300388787

En el otro caso,u=−9,623475382 = 3xno es posible obtener soluci´on. 4.

3x+1+ 3x−1−1 = 0 =⇒3·3x+3

x

3 −1 = 0 =⇒10·3

x₋_{3 = 0 =}_⇒

3x = 0,3 =⇒log 3x = log 0,3 =⇒xlog 3 = log 0,3 =⇒

x= log 0,3

log 3 =−1,095903274

1. log(x−1)−log(x+ 1) = 1−logx

2. 32x−1+ 3x+1−2 = 0 Soluci´on:

1.

log(x−1)−log(x+ 1) = 1−logx=⇒logx−1

x+ 1 = log 10

x x−1

x+ 1 = 10

x =⇒x

2₋₁₁_x₋_{10 = 0 =}_⇒

x= 11,84428877, x=−0,8442887702

De las dos soluciones hay una que no es posible ya que no existen logaritmos de n´umeros negativos, es decir, de las dos soluciones la ´

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2.

32x−1+ 3x+1−2 = 0 =⇒ 3

2x

3 + 3·3

x₋_{2 = 0 =}_⇒₃2x_{+ 9}_·₃x₋_{6 = 0}

u2+ 9u−6 = 0 =⇒u= 0,6234753829, u=−9,623475382

u= 0,6234753829 = 3x=⇒log 0,6234753829 = log 3x=⇒

xlog 3 = log 0,6234753829 =⇒

x= log 0,6234753829

log 3 =−0,4300388787

En el otro caso,u=−9,623475382 = 3xno es posible obtener soluci´on.

Problema 135 Resolver:

1. log(1 +x)−log(1−x) = 2

2. 32x−2·3x+ 1 = 0

Soluci´on:

1. log(1 +x)−log(1−x) = 2 =⇒log1 +x

1−x = log 100 =⇒

1 +x= 100(1−x) =⇒x= 99 101

2. 32x−2·3x+ 1 = 0 haciendot= 3x tenemos que:

t2 −2t+ 1 = 0 =⇒ t = 1, deshaciendo el cambio de variable tene-mos:

t= 3x = 1 =⇒x= 0

Problema 136 Resolver las siguientes ecuaciones

1. log(x2−2) + 1 = log(x+ 1) + log(x−1)

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1.

log(x2−2) + 1 = log(x+ 1) + log(x−1)

log(x2−2) + log 10 = log(x+ 1) + log(x−1)

log 10(x2−2) = log(x2−1) =⇒10x2−20 =x2−1 =⇒x=± √

19 9 La soluci´on negativa no es v´alida, ya que no existen logaritmos de

n´umeros negativos y, por tanto,x=

√

19 9 . 2.

32x−1+ 3x+1−1 = 0 =⇒ (3

x₎2

3 + 3·3

x₋_{1 = 0}

Si hacemost= 3x nos queda

t2

3 + 3t−1 = 0 =⇒t

2_{+ 9}_t₋_{3 = 0 =}_⇒_t_{= 0}_,_321825; _t₌₋₉_,₃₂₁₈₂₅

Deshaciendo el cambio de variable tendremos:

            

3x = 0,321825 =⇒log 3x= log 0,321825 =⇒xlog 3 = log 0,321825 =⇒

=⇒x= log 0,321825

log 3 =−1,03198

3x =−9,321825 no tiene soluci´on

Problema 137 Resolver:

1. log(5x+ 1)−logx= 1−log(1−x)

2. 22x−1−2x+1+ 2 = 0 3.

(

2x−1+ 3y+1 = 3 2x+1− 3y−1 = 1 Soluci´on:

1.

5x+ 1

x =

10

1−x =⇒5x

2_{+ 6}_x₋_{1 = 0 =}_⇒

x=−1,348331477, x= 0,1483314773 La soluci´on negativa no es v´alida.

2.

t2

2 −2t+ 2 = 0 =⇒t= 2 =⇒x= 1 3.

(

u/2+ 3v= 3 2u− v/3 = 1 =⇒

(

u= 24/37

v= 33/37 =⇒

(

x=−0,2712129366